Posted by **Lisa** on Monday, July 22, 2013 at 3:55pm.

Find all solutions in the interval 0 degrees<θ<360 degrees. If rounding necessary, round to the nearest tenth of a degree. 17sec2 θ − 15tanθsecθ − 15 = 0

- Trig -
**Steve**, Monday, July 22, 2013 at 4:52pm
17sec^2θ - 15secθ tanθ - 15 = 0

15secθ tanθ = 17sec^2θ - 15

225 sec^2θ tan^2θ = 289sec^4θ - 510sec^2θ + 225

225sec^4θ - 225sec^2θ = 289sec^4θ - 510sec^2θ + 225

64sec^4θ - 285sec^2θ + 225 = 0

That's just a quadratic in sec^2θ, so just solve it and you have your solution candidates.

However, because we squared things, there may be spurious solutions, so you have to check the values in the original equation.

In radians, I get .159,1.000,2.141,2.982

- Trig -
**Reiny**, Monday, July 22, 2013 at 5:01pm
17sec2 θ − 15tanθsecθ − 15 = 0

17/cos^2 Ø - 15(sinØ/cosØ)(1/cosØ) - 15 = 0

times cos^2 Ø

17 - 15sinØ - 15cos^2 Ø = 0

17 - 15sinØ - 15(1 - sin^2 Ø) = 0

15sin^2 Ø - 15sinØ + 2 = 0

sinØ = (15 ± √105)/30

sinØ = .158435 or sinØ = .841565

Ø = 9.1° or 170.9° or Ø = 57.3° or 122.7°

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